Question: Is ${242785}$ divisible by $9$ ?
A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {242785}= &&{2}\cdot100000+ \\&&{4}\cdot10000+ \\&&{2}\cdot1000+ \\&&{7}\cdot100+ \\&&{8}\cdot10+ \\&&{5}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {242785}= &&{2}(99999+1)+ \\&&{4}(9999+1)+ \\&&{2}(999+1)+ \\&&{7}(99+1)+ \\&&{8}(9+1)+ \\&&{5} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {242785}= &&\gray{2\cdot99999}+ \\&&\gray{4\cdot9999}+ \\&&\gray{2\cdot999}+ \\&&\gray{7\cdot99}+ \\&&\gray{8\cdot9}+ \\&& {2}+{4}+{2}+{7}+{8}+{5} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${242785}$ is divisible by $9$ if ${ 2}+{4}+{2}+{7}+{8}+{5}$ is divisible by $9$ Add the digits of ${242785}$ $ {2}+{4}+{2}+{7}+{8}+{5} = {28} $ If ${28}$ is divisible by $9$ , then ${242785}$ must also be divisible by $9$ ${28}$ is not divisible by $9$, therefore ${242785}$ must not be divisible by $9$.